Question

如何将此查询的结果与自身交叉加入？

sql-fiddle：http://sqlfiddle.com/#!2/88f5e2/2

SELECT t.tid ,count(*) AS count
FROM taxonomy_index t 
JOIN taxonomy_term_data td ON td.tid = t.tid
WHERE t.created > UNIX_TIMESTAMP() - 3 * 86400 AND td.vid = 1
GROUP BY t.tid
ORDER BY count DESC
LIMIT 0, 5;

查询结果为：

我需要：

4429    4429       6
2634    4429       5
1703    4429       4
1742    4429       4
4468    4429       4
4429    2634       .
2634    2634       .
1703    2634       .
1742    2634
4468    2634
4429    1703
2634    1703
1703    1703
1742    1703
4468    1703
....        ....

Answer 1

SELECT t_outer.tid as tid, t_inner.tid as tid_group, t_inner.count, t_inner.name
FROM taxonomy_index as t_outer
JOIN (
  SELECT t.tid ,count(*) AS count, td.name, td.vid
  FROM taxonomy_index t 
  JOIN taxonomy_term_data td ON td.tid = t.tid
  WHERE t.created > 0 AND td.vid = 1
  GROUP BY t.tid
) t_inner
WHERE t_outer.created > 0 AND t_inner.vid = 1
ORDER BY count DESC
LIMIT 0, 20;

的产率：

TID     TID_GROUP   COUNT   NAME
4429    4429        3       A1
4468    4429        3       A1
4429    4429        3       A1
2634    4429        3       A1
4429    4429        3       A1
2634    4429        3       A1
4429    2634        2       B1
4468    2634        2       B1
4429    2634        2       B1
2634    2634        2       B1
4429    2634        2       B1
2634    2634        2       B1
2634    4468        1       C1
4429    4468        1       C1
4468    4468        1       C1
4429    4468        1       C1
2634    4468        1       C1
4429    4468        1       C1

http://sqlfiddle.com/#!2/88f5e2/29

信用到期的信用：count without group

更新：对不起，忘了约束外部数据。更新2：我也忘记了tid列表

复杂查询中的CROSS JOIN

1 个答案: